Answer:
C) 30,000
Explanation:
According to the given information, the bacterial mRNA consists of about 800 nucleotides. Three consecutive nucleotides together make one genetic codon which in turn codes for one specific amino acid in the protein encoded by this mRNA.
So, an mRNA with 800 nucleotides will have total 800/3 = 266.67 or 266 genetic codes. The protein encoded by this mRNA would have a total of 266 amino acids.
Given that one amino acid imparts 110 units to the molecular weight of the protein, the protein with 266 amino acids have the molecular weight= 266 x 110 = 29260, that is about 30,000.
Proteins in a protein complex are linked by non - covalent protein - protein interactions. The process of complex formation comprises of steps namely : ... An encounter complex is formed that either proceeds towards final complex or dissociates again.
Answer:
Throughout the experiment at hand, the student will likely find that root space does indeed effect the length to which plantswill grow. Although I could not locate the table online, I will offer a general hypothesis for what the experiment will conclude. The student is testing the effects of root space on the way plants grow. The students experiment is well designed, keeping many variables constant and making sure that the amount of root spaceis the only difference between the plantsensures the accuracy of the experiment. Since we know that the experiment is well designed we are able to directly infer the effects of root space limitations without having to account for other factors. We can assume that the limited root space will in turn limit the extent to which a plant can grow. This is because plants need larger and deeper roots to support extended growth. Therefore, the plants in group Bwill grow smaller than those in group A.
Explanation:
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Correct question:
if a nondisjunction occurs at anaphase I of the first meiotic division, what will the proportion of abnormal gametes (for the chromosomes involved in the nondisjunction)?
Answer:
100%
Explanation:
Nondisjunction at meiosis-I means that two homologous chromosomes of at least one homologous pair fail to separate from each other during anaphase-I. This would result in the formation of one cell with one extra chromosome and the other with one less chromosome by the end of meiosis-I. Meiosis-II in these two cells would maintain this chromosome number in the daughter cells. Therefore, out of the total four gametes formed by the end of the meiosis, two would have one extra chromosome and would be denoted as "n+1". The rest of the two gametes would have one less chromosome and would be denoted as "n-1".
Answer:
Properties which are needed for something to be considered as living are -
I) Movement
ii) Growth
iii) Respiration
iv) Reproduction
v) And other life processes such as ( transportation, excreting etc)
